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Problem 1:

If $n$ people are seated in a random manner in a row containing $2n$ seats, what is the probability that no two people will occupy adjacent seats?

I know that the probability that $n$ people are seated in $2n$ rows is ${2n\choose n}$

I also know that the answer to this problem is $n+1\over {2n\choose n}$

Why does $n+1$ solidify that the people will not be sitting next to each other? I somewhat understand it, but I would appreciate a more logical explanation.

Problem 2:

What is the probability that a group of $50$ senators selected at random will contain one senator from each state?

So obviously the set of $50$ senators is the combinatorial ${100\choose 50}$ and we want $1$ senator from each state so that's ${2\choose 1}$.

So we have ${2\choose 1}\over {100\choose 50}$, but the ${2\choose 1}$ should be ${2\choose 1}^{50}$ for all 50 states. Why do we have to put the numerator to the 50th power? Shouldn't ${2\choose 1}$ suffice that each senator should be from $1$ state?

Thanks

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2 Answers

up vote 3 down vote accepted

ad 1: $2n\choose n$ is the number of ways to seat $n$ people without regarding their identities. Now, how many of these will have no two people adjacent? You could occupy all odd seats, leaving the even ones free. Note that this will also leave seat $2n$ free, although this is not necessary in order to "keep the distance". So, you have one additional unoccupied seat which you can move about. It can be to the left of any person ($n$ possibilities), or to the right of the rightmost person ($1$ possibility). Therefore, you have $n+1$ ways to seat $n$ people.

ad 2: You can select senators state by state. For the first state, there are $2\choose 1$ possibilities to select exactly one senator. Likewise for the second state, and so on.

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Wow! That makes perfect sense, thanks! Any ideas on the second problem? –  switz Feb 9 '12 at 11:06
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Alternate wording of same argument: There are $n+1$ people, the $n$ students in the class plus the teacher. There is only $1$ way to seat them so no two are adjacent. Now the teacher leaves. The seat she used to occupy can be any one of the $n+1$ seats that were used. –  André Nicolas Feb 9 '12 at 16:15
    
@AndréNicolas I like that one because it's more symmetrical than my argument. –  Ansgar Esztermann Feb 10 '12 at 9:00
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For problem 2, why not consider the smaller problem where there are only two states? Small enough that you can write down all the possibilities and see what the answer is, and then compare to the formulas you are asking about.

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